Do cosine, tangent, and cotangent have similar limit identities to sine?

[MathematicalPhysicist]

i am familiar to the this limit:
$$\lim_{x\rightarrow 0} sin(x)/x=1$$

my question is does cosine,tg and cot have a similar idnetity limit or it's only a special case only in sine function?

 

[Muzza]

$$\lim_{x\rightarrow 0} \frac{cosx - 1}{x}=0$$ perhaps...

 

[MathematicalPhysicist]

Originally posted by Muzza
$$\lim_{x\rightarrow 0} \frac{cosx - 1}{x}=0$$ perhaps...

i would think that the proof to what you have give is from the fact that cos(0)=1 which is simple but the proof to sin(x)/x is a little bit more complex (or at least to proof i read from my book).

 

[matt grime]

Both proofs are the same 'complexity' - l'Hopital's rule in both cases. (or, less slickly, Taylor series, which is the same thing really)



that cos(0)=1 does not imply

lim (cos(x)-1)/x is zero,

or surely it would also imply

lim (cos(x)-1)/x^2

tends to zero, when it doesn't. It would be instructive for you to find the limit.

 

[Hurkyl]

And the other trig functions are just a ratio of sines, cosines, and/or 1, and can be found from the limits of these.

 

[Muzza]

Originally posted by loop quantum gravity
i would think that the proof to what you have give is from the fact that cos(0)=1 which is simple but the proof to sin(x)/x is a little bit more complex (or at least to proof i read from my book).

Did you miss the x in the denominator?

 

[MathematicalPhysicist]

Originally posted by matt grime
Both proofs are the same 'complexity' - l'Hopital's rule in both cases. (or, less slickly, Taylor series, which is the same thing really)



that cos(0)=1 does not imply

lim (cos(x)-1)/x is zero,

or surely it would also imply

lim (cos(x)-1)/x^2

tends to zero, when it doesn't. It would be instructive for you to find the limit.

it seems you didnt notice i said the proof i read from my book which is more a geometric proof of sin(x)/x than a series proof or by lohiptal's rule and this proof (i think) is more complex than the ways you have listed.


i wish i could give you the proof but the problem is the book is in hebrew and my scanner doesn't work.

 

[matt grime]

I saw that you wrote 'in my book' or similar, but you didn't explain what the proof was in your book, or why you thought that the fact that cos(0)=1 was sufficient to imply the other identity, so I had to guess what you considered to be the method of proof. My proofs of both have exactly the same level of theory and computation behind them.

Explain the idea behind the geometric proof (I can't think of a rigorous one right now).

Edit: and this is the calculus forum isn't it?

 

[MathematicalPhysicist]

the proof uses the unit circle (where the radius is 1), here's a similar proof to what i read in the book:
http://www.geocities.com/pkving4math2tor4/4_the_elem_transc_func/4_01_01_03_lim_of_trig_func.htm

it's in section 3.

now my follow up question does exist such a proof (with the use of unit circle) for cosx-1/x?


btw i saw the solution also to that in the above link but it's not a geometric solution.

 

[matt grime]

To my mind that is a far *simpler* proof than invoking l'Hopital, and might even count as one for 'the book', it's just that the complexity in l'Hopital is hidden. Here we don't even assume sin is differentiable at 0

As for whether one exists for the cosine example, well, you'd have to play around with it.

 

[vadik]

Proof of (cos x - 1) / x = 0

$$\lim_{x\rightarrow 0} \frac{cosx - 1}{x}=0$$ perhaps...

The proof of this isn't too difficult. Let's express cos x as a taylor series:
cos x = sum n=0 to inf. x^2n(-1)^n/(2n)! , which is the same as
= 1 + sum n=1 to inf. x^2n(-1)^n/(2n)! thus
cosx - 1 = sum n=1 to inf. x^2n(-1)^n/(2n)! and dividing by x we get
(cos x - 1)/x = sum n=1 to inf. x^(2n-1)(-1)^n/(2n)!, which is defined for all x and also to x = 0 and of course
sum n=1 to inf. 0^(2n-1)(-1)^n/(2n)! = 0

The proof for sin x/x is of the same kind, and not more difficult.

P.S. I've never used this program before and I don't know how to mark signs, but I think you got clear of what I wrote.

 

[DrMatrix]

Both proofs are the same 'complexity' - l'Hopital's rule in both cases. (or, less slickly, Taylor series, which is the same thing really)

Um . . . In order to use l'Hopital's rule to evaluate $\lim_{x\rightarrow 0} sin(x)/x$, you need to take the derivative of sin(x), but if I recall corectly, the usual proof that d sin(x)/dx = cos(x) uses $\lim_{x\rightarrow 0} sin(x)/x=1$.

edited to add

Similarly, getting the Taylor series also requires us to take derivatives of sin(x).

 

[matt grime]

There are other ways of determining the limits as the whole of the question and its answers tells us.

 

[vadik]

It is possible (and easy) to prove dsinx/dx = cosx WITHOUT using limit lim sinx/x=0. A have made a picture to show it (it goes using geometry) but I don't know how to show it here.

 

[vadik]

Oh yeah, I learned how to attach:) So here is the prove that doesn't use the fact sinx/x-->1 when x--> 0 nor (cosx-1)/x-->0, when x-->0. (if i did the attach properly)

 

[DrMatrix]

You can't just say that the less delta-x is, the closer l is to a straight line. (It is true, but not established.) In order to show that, you need to know sinx/x --> 1.

Edit: I had need not know. Changed to need to know.

 

[vadik]

ups

Yea, true, I didn't see it. But from the picture above you can see, that
xcosx < sinx < x, which implies, that cosx < sinx/x < 1 and taking limits we see: lim_x-->0 cos x <= lim_x-->0 sinx/x <= 1, hence lim x-->0 sinx/x = 0.

The first inequality xcosx < sinx follows from the fact that the area of the sector OCD (xcosxcosx/2) is smaller than area of the triangle OCB (cosxsinx/2). cosx <> 0. That is xcosxcosx/2 < cosxsinx/2 <=> xcosx < sinx. And sinx < x we get the same way.

 

[HallsofIvy]

Yea, true, I didn't see it. But from the picture above you can see, that
xcosx < sinx < x, which implies, that cosx < sinx/x < 1 and taking limits we see: lim_x-->0 cos x <= lim_x-->0 sinx/x <= 1, hence lim x-->0 sinx/x = 0.

? This shows that lim_x-->0 sinx/x= 1, not 0.

 

[vadik]

ahh.. yes, my point was to show, that it is 1. Writing 0 was just a mistake . Sorry.

 

[Abcopp]

some friends of mine came up with something similar to what all of you are talking about:
lim of tan z(x) / sin n(x) = z/n

the limit is as x approaches 0

 

[Elucidus]

Apologies in advance if this is obvious but here are several related results without L'Hospital's Rule or Taylor Series...

(1) $$\lim_{x \rightarrow 0} \frac{\cos x -1}{x} = 0.$$

Proof.
Since x is approaching 0 it is safe to assume $-\pi / 2 < x < \pi / 2$. For x in this interval, cos(x) > 0 and therefore

$$\cos x = \sqrt{1-\sin^2 x}$$​

So,

$$\lim_{x \rightarrow 0} \frac{\cos x -1}{x} = \lim_{x \rightarrow 0} \frac{\sqrt{1 - \sin^2 x} -1}{x} = \lim_{x \rightarrow 0} \frac{\sqrt{1 - \sin^2 x} -1}{x} \cdot \frac{\sqrt{1-\sin^2 x} + 1}{\sqrt{1-\sin^2 x} + 1}= \lim_{x \rightarrow 0} \frac{-\sin^2 x}{x \sqrt{1-\sin^2 x} + 1}$$

$$=\lim_{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{-\sin x}{\sqrt{1-\sin^2 x}+1}= (1) \cdot \frac{0}{\sqrt{1-0} + 1} = 0.$$


(2) $$\lim_{x \rightarrow 0} \frac{\sin \alpha x}{x} = \lim_{x \rightarrow 0} \frac{\tan \alpha x}{x} = \alpha$$


Proof.(For sine)
$$\lim_{x \rightarrow 0} \frac{\sin \alpha x}{x} = \lim_{x \rightarrow 0} \left( \frac{\sin \alpha x}{\alpha x} \cdot \frac{\alpha x}{x} \right) = (1)(\alpha) = \alpha.$$


(3) $$\lim_{x \rightarrow 0} \frac{\sin \alpha x}{\sin \beta x} = \lim_{x \rightarrow 0} \frac{\tan \alpha x}{\tan \beta x} = \frac{\alpha}{\beta}$$

Proof.(For sine)
$$\lim_{x \rightarrow 0} \frac{\sin \alpha x}{\sin \beta x} = \lim_{x \rightarrow 0} \left( \frac{\sin \alpha x}{\alpha x} \cdot \frac{\alpha x}{\beta x} \cdot \frac{\beta x}{\sin \beta x} \right) = (1)(\alpha / \beta)(1) = \frac{\alpha}{\beta}$$

--Elucidus